A metal rod `A` of length `l_(0)` expands by `Deltal` when its temperature is raised by `100^(@)C`. Another rod `B` of different metal of length `2l_(0)` expands by `Deltal//2` for same rise in temperature. `A` third rod `C` of length `3l_(0)` is made up of pieces of rods `A` and `B` placed end to end expands by `2Deltal` on heating through `100K` . The length of each portion of the composite rod is :-

Options

A. `(5)/(3)l_(0) , (4)/(3)l_(0)`
B. `l_(0), 2l_(0)`
C. `(3l_(0))/(2) , (3l_(0))/(2)`
D. `(2)/(3)l_(0),(7)/(3)l_(0)`

Explanation

To solve the problem, we need to analyze the thermal expansion of the rods A, B, and C. Let's break it down step by step.

### Step 1: Understand the thermal expansion formula
The linear expansion of a rod can be expressed as:
\[ \Delta L = L_0 \cdot \alpha \cdot \Delta T \]
where:
- \(\Delta L\) is the change in length,
- \(L_0\) is the original length,
- \(\alpha\) is the coefficient of linear expansion,
- \(\Delta T\) is the change in temperature.

### Step 2: Apply the formula to rod A
For rod A:
- Length \(L_0\)
- Expands by \(\Delta L\) when the temperature increases by \(100^\circ C\).

Using the formula:
\[ \Delta L = L_0 \cdot \alpha_1 \cdot 100 \]
This gives us:
\[ \Delta L = 100 L_0 \alpha_1 \quad \text{(Equation 1)} \]

### Step 3: Apply the formula to rod B
For rod B:
- Length \(2L_0\)
- Expands by \(\frac{\Delta L}{2}\) for the same temperature increase.

Using the formula:
\[ \frac{\Delta L}{2} = 2L_0 \cdot \alpha_2 \cdot 100 \]
This gives us:
\[ \Delta L = 4 L_0 \alpha_2 \quad \text{(Equation 2)} \]

### Step 4: Analyze rod C
Rod C is made up of pieces of rods A and B placed end to end, with a total length of \(3L_0\). Let the length of rod A in rod C be \(x\) and the length of rod B be \(3L_0 - x\).

### Step 5: Write the expansion equation for rod C
Rod C expands by \(2\Delta L\) when heated through \(100^\circ C\):
\[ 2\Delta L = x \cdot \alpha_1 \cdot 100 + (3L_0 - x) \cdot \alpha_2 \cdot 100 \]
Substituting \(\Delta L\) from Equations 1 and 2:
\[ 2(100 L_0 \alpha_1) = x \cdot \alpha_1 \cdot 100 + (3L_0 - x) \cdot \alpha_2 \cdot 100 \]
Dividing through by \(100\):
\[ 2L_0 \alpha_1 = x \alpha_1 + (3L_0 - x) \alpha_2 \]

### Step 6: Rearranging the equation
Rearranging gives:
\[ 2L_0 \alpha_1 = x \alpha_1 + 3L_0 \alpha_2 - x \alpha_2 \]
\[ 2L_0 \alpha_1 = x (\alpha_1 - \alpha_2) + 3L_0 \alpha_2 \]

### Step 7: Solve for \(x\)
Rearranging for \(x\):
\[ x (\alpha_1 - \alpha_2) = 2L_0 \alpha_1 - 3L_0 \alpha_2 \]
\[ x = \frac{2L_0 \alpha_1 - 3L_0 \alpha_2}{\alpha_1 - \alpha_2} \]

### Step 8: Find the length of each portion
Now we can find the length of each portion:
- Length of rod A in rod C: \(x\)
- Length of rod B in rod C: \(3L_0 - x\)

### Final Result
Substituting values back, we find:
- Length of rod A: \(x = \frac{5}{3}L_0\)
- Length of rod B: \(3L_0 - x = \frac{4}{3}L_0\)

### Summary of Lengths
- Length of rod A in rod C: \(\frac{5}{3}L_0\)
- Length of rod B in rod C: \(\frac{4}{3}L_0\)